Two identical rods of metal with the same cross section are welded end to end, a

Question

Two identical rods of metal with the same cross section are welded end to end, as shown in figure a, with a temperature of T_c on the right side and T_h on the left side. (a) Derive an expression (don't just write it down!) for the total heat transfer rate (power transfer) between two surfaces at temperatures T_h and T_c, (T_h > T_c) across the two rods of metal welded end to end in series (shown in figure a), where the first rod has thermal conductivity k_1 and the second has thermal conductivity k_2. Both rods have the same length, L. (b) Suppose T_c = 0 degree C and T_h = 100 degree C. When the rods are in the configuration shown in figure a, 10 J is conducted at a constant rate from the left side to the right side in 2.0 min. How much time would be required to conduct 10 J if the rods were welded side to side between the same reservoirs, as in figure b?

Explanation / Answer

(a) dQ/dt = kA(T)/L

Let 'T' be the temperature at the point of welding

dQ/dt = k1A(Th-T)/L= k2A(T-Tc)/L

=> k1A(Th-T)/L= k2A(T-Tc)/L

=> k1Th + k2Tc = (k1 + k2)T

=> T = (k1Th + k2Tc) / (k1 + k2)

dQ/dt = k1A(Th-T)/L

=> dQ/dt = (k1A/L)[Th-  (k1Th + k2Tc) / (k1 + k2)]

=> dQ/dt = (k1A/L)[(k2Th - k2Tc) / (k1 + k2)]

=> dQ/dt = (k1k2A/L)[(Th - Tc) / (k1 + k2)]

=> dQ/dt = [k1k2/(k1 + k2)]*A*[(Th - Tc)/L]

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(b)

(dQ/dt) = (dQ/dt)1 + (dQ/dt)2

=> (dQ/dt) = (k1A/L)(Th-Tc) + (k2A/L)(Th-Tc)

=> (dQ/dt) = [(k1+ k2)A/L](Th-Tc)

(dQ/dt)a = (kAa/La)(TH - TC)

(dQ/dt)b = (kAb/Lb)(TH - TC)

=> (dQ/dt)b = (Ab/Aa)(La/Lb)(dQ/dt)a

=> (dQ/dt)b = (2)(2)(dQ/dt)a

=> (dQ/dt)b = 4*(10/2) = 20 J/min

Time taken to conduct 10J = 0.5 min = 30 s

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